## anonymous one year ago can't understand why integral of xe^((x^2)/2) is e^((x^2)/2) can someone explain how to calculate this integral please?

1. UsukiDoll

$\LARGE xe^{\frac{x^2}{2}}$ this one?

2. UsukiDoll

do you know u-substitution?

3. UsukiDoll

The u-substitution is an integration method that essentially involves using the chain rule in reverse.

4. UsukiDoll

hmm let's try rewriting that fraction $\LARGE xe^{\frac{1}{2}x^2}$ so we can see the exponent clearly.

5. UsukiDoll

we need to pick a u so that when we take that derivative of u we can substitute.. so what will be our u? Well we do see an x in front of the e.. Maybe we can have $\LARGE u=\frac{1}{2}x^2$

6. UsukiDoll

so what is the derivative of $\large \frac{1}{2}x^2 ?$

7. UsukiDoll

we can use product rule to make this a bit easier. Let f(x) = 1/2 and g(x) = x^2 recall that the product rule is f(x)g'(x) +f'(x)g(x)

8. UsukiDoll

so what is the derivative of f(x) = 1/2 ?

9. UsukiDoll

similarly what is the derivative of $\large g(x) = x^2$

10. anonymous

so if $u= (1/2)x^2 -> u' = x$

11. UsukiDoll

yeah but we have to write this differently it will be du = x dx

12. anonymous

so now we have $xe^u dx$

13. UsukiDoll

hmm how do I explain this.... we just have $e^u$ and substitute the u back .. don't forget to add +C

14. UsukiDoll

remember du = x dx ? since there's nothing on the left hand side.... oh man .. I'm having trouble explaining this part. . . usually when there's nothing ... it's just blank like 1 sorry I'm in panic mode I can't figure out how to type this well

15. UsukiDoll

but we just have $\LARGE e^u +C$ now substitute $\LARGE u=\frac{1}{2}x^2$ back into $\LARGE e^u +C$

16. anonymous

ok, so $xe^u dx -> xe^u (1/x) dy -> e^u du$ (+c...)

17. anonymous

du

18. UsukiDoll

you don't have to write dx you don't have to write du

19. UsukiDoll

du = x dx it's just .... @ganeshie8 can you help me explain this? I know what's going on just can't put it into words

20. anonymous

Ok, I think I understand, thank you very much!

21. UsukiDoll

but in the end you should have $\LARGE e^u +C$ now plug in u back into this equation

22. UsukiDoll

and once we have our antiderivative, we can simply check by taking the derivative.

23. UsukiDoll

you should have something like $\LARGE e^{\frac{1}{2}x^2} +C$

24. UsukiDoll

now take the derivative of that equation...

25. UsukiDoll

that's how to check :)

26. anonymous

so we get back to $e^((x^2)/2)$

27. anonymous

yes, +c

28. UsukiDoll

mhm

29. UsukiDoll

if you take the derivative of your antiderivative, we should be getting that x in front of that e

30. anonymous

Yes, I wrote the integral

31. UsukiDoll

$\LARGE f(x) = e^{\frac{1}{2}x^2} +C$ $\LARGE f'(x) = xe^{\frac{1}{2}x^2} +C$ because the derivative is $\frac{1}{2}x^2 \rightarrow \frac{1}{2} (2x) + 0(x^2) \rightarrow x$ and that's placed in front of the e.

32. UsukiDoll

the derivative of any e^x is always in the front and the exponent part of e^x just stays there :)

33. anonymous

Excellent. Thanks a lot!